I need to create a java function that will only run for 30 minutes, and at the end of the 30 minutes it executes something. But it should also be able to self terminate before the given time if the right conditions are met. I don't want the function to be sleeping as it should be collecting data, so no sleeping threads.
Thanks
Use: Timer.schedule( TimerTask, long )
public void someFunctino() {
// set the timeout
// this will stop this function in 30 minutes
long in30Minutes = 30 * 60 * 1000;
Timer timer = new Timer();
timer.schedule( new TimerTask(){
public void run() {
if( conditionsAreMet() ) {
System.exit(0);
}
}
}, in30Minutes );
// do the work...
.... work for n time, it would be stoped in 30 minutes at most
... code code code
}
Get the start time with System.currentTimeMillis(), calculate the time when to stop and check the current time every now and then while you're collecting the data you want to collect. Another way would be to decouple the timer and the data collecting, so that each of them could run in their own threads.
For a more specific answer, it would be helpful if you would tell what data you are collecting and how you are collecting it.
Something like this will work:
long startTime = System.currentTimeMillis();
long maxDurationInMilliseconds = 30 * 60 * 1000;
while (System.currentTimeMillis() < startTime + maxDurationInMilliseconds) {
// carry on running - 30 minutes hasn't elapsed yet
if (someOtherConditionIsMet) {
// stop running early
break;
}
}
The modern java.util.concurrent way would be using ExecutorService. There are several invoke methods taking a timeout.
Here's a kickoff example:
public static void main(String args[]) throws Exception {
ExecutorService executor = Executors.newSingleThreadExecutor();
executor.invokeAll(Arrays.asList(new Task()), 30, TimeUnit.MINUTES);
executor.shutdown();
}
where Task look like this:
public class Task implements Callable<String> {
#Override
public String call() {
// Just a dummy long running task.
BigInteger i = new BigInteger("0");
for (long l = 0; l < Long.MAX_VALUE; l++) {
i.multiply(new BigInteger(String.valueOf(l)));
// You need to check this regularly..
if (Thread.interrupted()) {
System.out.println("Task interrupted!");
break; // ..and stop the task whenever Thread is interrupted.
}
}
return null; // Or whatever you'd like to use as return value.
}
}
See also:
Lesson: Concurrency
Related
I managed to get the customers showing up at random times, the only problem I have now is wrapping the whole thing in a 2-minute timer without conflicting with the customer arrivals.
static Timer timer = new Timer();
static class Task extends TimerTask {
#Override
public void run() {
int delay = (new Random().nextInt(5)+2) * 1000;
timer.schedule(new Task(), delay);
System.out.println("Hi " + delay);
}
}
public static void main(String args[]) {
new Task().run();
}
This is what I have that works so far. I've tried wrapping it in another TimerTask, and using System.nanoTime() and a while loop. They both just end up conflicting with the customer arrivals. I usually like to figure things out on my own, but I've been working on this one part for hours and I can't figure it out.
Why not to use a sleep with a do-while loop like:
long durationMillis = TimeUnit.MINUTES.toMillis(2);
long endTime = System.currentTimeMillis() + durationMillis;
do {
int delaySec = ThreadLocalRandom.current().nextInt(2, 7);
TimeUnit.SECONDS.sleep(delaySec);
System.out.println("Hi " + delaySec);
} while (System.currentTimeMillis() < endTime);
Change delaySec semantics if you don't need to wait for whole seconds or 6 second wait is not inclusive.
I want to Schedule a job to run at multiple of 2 seconds, which is 2,4,8,16,32 seconds. Second fire should happen after two seconds of completion of first fire, Third fire should happen after 4 seconds of completion of second fire and so on. The next fire is based on status we get from previous fire, based on which it will be decided whether we need to trigger next fire or not.
Can somebody tell me how can I use quartz scheduler to achieve this?
If I use SimpleTrigger.withIntervalInSeconds(2) it runs a job after every 2 seconds where as I want time interval should be increased with multiple of 2 in every firing.
Perhaps you could forget trying to set up a single trigger, but use multiple triggers. My Java is not good in this area, so this is in pseudocode:
delay = 2
repeat
TriggerOnceAfter(delay)
delay <- delay * 2
WaitUntilTriggered()
until (finished)
I am not sure how to implement the WaitUntilTriggered() method; you my need to add a signalling flag to the triggered code for WaitUntilTriggered() to look at.
That will give delays of 2, 4, 8, ...
This is a simplified implementation that will invoke a Runnable at the requested schedule:
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.TimeUnit;
public class Tasker {
private int numberOfRuns; //how many times job executed
private int timeBetweenRuns;//seconds
Tasker(int numberOfRuns){
this.numberOfRuns = numberOfRuns;
timeBetweenRuns = 2;
execute();
}
private void execute() {
for (int counter = 0; counter < numberOfRuns ; counter++) {
CountDownLatch latch = new CountDownLatch(1);
Job job = new Job(latch, timeBetweenRuns);
job.run();
try {
latch.await();
TimeUnit.SECONDS.sleep(timeBetweenRuns);
} catch (InterruptedException ex) {
ex.printStackTrace();
}
timeBetweenRuns *=2;
}
}
public static void main(String[] args){
new Tasker(5);
}
}
class Job implements Runnable {
private int seconds;
private CountDownLatch latch ;
Job(CountDownLatch latch , int seconds){
this.latch = latch;
this.seconds = seconds;
}
#Override
public void run() {
System.out.println("Job runs "+ seconds +" after previous one");
latch.countDown();
}
}
I want to run java code for a certain duration ,say 16 hours! I have a java code that runs for approximately an hour.I want this to run repeatedly for 16 hours. So I have a parameter that is passed by the user through Jenkins ! I access this value using
System.getenv("Duration");
Now, I want to exit the execution after the specified time. So suppose the user selected 16, the script should run for 16 hours and then exit.
Accepting input from Jenkins user as shown in the image
I saw some other questions, but most of them were dealing with timers for either few seconds or few minutes. I need an efficient solution. Thanks :)
FYI - Environment - Jenkins+TestNG+Maven+Java
EDIT :
long start = System.currentTimeMillis();
long end = start + durationInHours*60*60*1000;
while (System.currentTimeMillis() < end)
{
//My code here runs for approx. 50 mins!
}
Now suppose the user chooses the value 3 hours, I want the while loop to exit after 3 hours. But this does not happen as it has not yet completed 3 hours when checking the while condition.So it enters the while condition even the 4th time(since time elapsed is 150 mins which is less than 180 mins) it ends after 3 hours ten mins.
How to make it exit the while loop as soon as 180 mins is reached ?
P.S - I could do the math first,( iterations =durationFromUser/codeDuration) and then run a for loop, but I don't want to do this as my script length may vary.
EDIT 2:
boolean alive = true;
Timer timer = new Timer();
#Test() //Annotation from TestNG
public void public void jenkinsEntryPoint()
{
String duration = System.getenv("Duration");
int durationInHours=Integer.parseInt(duration);
long end = System.currentTimeMillis() + durationInHours*60*60*1000;
TimerTask task = new TimerTask() {
public void run() {
alive = false;
};
timer.schedule(task, end);
while (alive) {
//My code here runs for approx. 50 mins!
function1();
}
}
void function1() {
function2();
}
private void function2() {
for(i=0;i<8;i++)
{
while(alive)
{
//long running code
sleep(1000);
//Some more code
sleep(2000);
//Some more code
//Suppose time elapses here, I want it to quit
//But its continuing to execute
.
.
.
.
}
}
}
The while condition will only be evaluated between script invocations (as you've seen). You will have to break out of your long running from inside of it.
I would typically use a Timer to set a "global" boolean that you would check from inside the loops in your long running code.
Something like this. Notice checks against 'alive' would have to be in all you long loops...
boolean alive = true;
Timer timer = new Timer();
public void jenkinsEntryPoint()
long end = System.currentTimeMillis() + durationInHours*60*60*1000;
TimerTask task = new TimerTask() {
public void run() {
alive = false;
};
timer.schedule(task, end);
while (alive) {
//My code here runs for approx. 50 mins!
yourLongRunningCode()
}
public void yourLongRunningCode() {
while (alive) {
doStuff();
}
}
I tried ScheduledThreadPoolExecutor and it worked!
ScheduledThreadPoolExecutor exec = new ScheduledThreadPoolExecutor(1);
exec.scheduleAtFixedRate(new Runnable() {
public void run() {
System.out.println("Time's Up According To ScheduledThreadPool");
alive = false;
}
}, durationInHours, 1, TimeUnit.HOURS);
This function will be executed after "durationInHours".
Thanks #TedBigham :)
What is the difference between these 2 methods of Timer class :
schedule(TimerTask task, long delay, long period)
and
scheduleAtFixedRate(TimerTask task, long delay, long period)
Documentation doesn't make the difference between them clear.
The documentation does explain the difference:
schedule:
In fixed-delay execution, each execution is scheduled relative to the actual execution time of the previous execution. If an execution is delayed for any reason (such as garbage collection or other background activity), subsequent executions will be delayed as well.
So, suppose the delay is 5 seconds, and each task takes 2 seconds, you would get
TTWWWTTWWWTTWWWTT
where T means 1 second for the task execution, and W means 1 second waiting.
But now suppose that a long GC (represented by a G) happens and delays the second task, the third one will start 5 seconds after the start of the second one, as if the long GC didn't happen:
TTWWWGGTTWWWTTWWWTT
The third task starts 5 seconds after the second one.
scheduleAtFixedRate:
In fixed-rate execution, each execution is scheduled relative to the scheduled execution time of the initial execution. If an execution is delayed for any reason (such as garbage collection or other background activity), two or more executions will occur in rapid succession to "catch up.".
So, with the same delay as above, and the same GC, you would get
TTWWWGGTTWTTWWWTT
The third task task starts 3 seconds instead of 5 after the second one, to catch up.
Thanks #Nizet's answer, I have written a sample code for some people who want to practice and learn.
import java.util.Timer;
import java.util.TimerTask;
public class TimerTest {
public static void main(String args[]){
TimerTest.DelayTask task = new DelayTask();
Timer timer = new Timer();
/**
* Use schedule or scheduletAtFixedrate and check the printed result
*/
timer.schedule(task, 0, 5000);
//timer.scheduleAtFixedRate(task, 0, 5000);
}
public static boolean stop = false;
public static void delayOneSec(String status){
try{
System.out.print(status);
Thread.sleep(1000);
}catch(Exception e){
e.printStackTrace();
}
}
static class DelayTask extends TimerTask{
int count = 2;
#Override
public void run() {
// TODO Auto-generated method stub
stop = true;
for(int i = 0; i < count; i++){
TimerTest.delayOneSec("T");
}
if(count == 2){
count = 6;
}else{
count = 2;
}
stop = false;
new PrintW().start();
}
}
static class PrintW extends Thread{
#Override
public void run(){
while(!stop){
TimerTest.delayOneSec("W");
}
}
}
}
The task itself will repeat to take 2 seconds or 6 seconds. Let's see the result of each scenario.
When using timer.schedule(task, 0, 5000);, the output is TTWWWTTTTTTTTWWWTTTTTTTTWWWTTTTTTTT. As you can see, the timer follow the rules like below, wait till period time outs if task finishes in time, launch next task immediately if current task lasts more than period.
When using timer.scheduleAtFixedRate(task, 0, 5000);, the output is TTWWWTTTTTTTTWWTTTTTTTTWWTTTTTTTTWWTTTTTTTTWWTTTTTTTTWWTTTTTTTT. Things are a little different now. The javadoc
two or more executions will occur in rapid succession to "catch up."
takes effect here. As you can see, ignoring the first TTWWW, every two tasks will print TTTTTTTTWW and it lasts 10 seconds(two periods).
Let's dig into the source code of Timer.
public void schedule(TimerTask task, Date firstTime, long period) {
if (period <= 0)
throw new IllegalArgumentException("Non-positive period.");
sched(task, firstTime.getTime(), -period);
}
public void scheduleAtFixedRate(TimerTask task, long delay, long period) {
if (delay < 0)
throw new IllegalArgumentException("Negative delay.");
if (period <= 0)
throw new IllegalArgumentException("Non-positive period.");
sched(task, System.currentTimeMillis()+delay, period);
}
As you can see, the period is transferred to negative value in schedule method. Let's see what's the difference when scheduling it.
The below code is in the mainloop of TimerThread,
currentTime = System.currentTimeMillis();
executionTime = task.nextExecutionTime;
if (taskFired = (executionTime<=currentTime)) {
if (task.period == 0) { // Non-repeating, remove
queue.removeMin();
task.state = TimerTask.EXECUTED;
} else { // Repeating task, reschedule
queue.rescheduleMin(
task.period<0 ? currentTime - task.period
: executionTime + task.period);
}
}
}
It's where magic happens, for schedule method, the next task execution time is based on the currentTime which is calculated right before the this task runs. That means, every task's execution time only be related with previous task starts time.
In the code, the variable timer would specify the duration after which to end the while loop, 60 sec for example.
while(timer) {
//run
//terminate after 60 sec
}
long start = System.currentTimeMillis();
long end = start + 60*1000; // 60 seconds * 1000 ms/sec
while (System.currentTimeMillis() < end)
{
// run
}
you should try the new Java Executor Services.
http://docs.oracle.com/javase/6/docs/api/java/util/concurrent/ExecutorService.html
With this you don't need to program the loop the time measuring by yourself.
public class Starter {
public static void main(final String[] args) {
final ExecutorService service = Executors.newSingleThreadExecutor();
try {
final Future<Object> f = service.submit(() -> {
// Do you long running calculation here
Thread.sleep(1337); // Simulate some delay
return "42";
});
System.out.println(f.get(1, TimeUnit.SECONDS));
} catch (final TimeoutException e) {
System.err.println("Calculation took to long");
} catch (final Exception e) {
throw new RuntimeException(e);
} finally {
service.shutdown();
}
}
}
If you can't go over your time limit (it's a hard limit) then a thread is your best bet. You can use a loop to terminate the thread once you get to the time threshold. Whatever is going on in that thread at the time can be interrupted, allowing calculations to stop almost instantly. Here is an example:
Thread t = new Thread(myRunnable); // myRunnable does your calculations
long startTime = System.currentTimeMillis();
long endTime = startTime + 60000L;
t.start(); // Kick off calculations
while (System.currentTimeMillis() < endTime) {
// Still within time theshold, wait a little longer
try {
Thread.sleep(500L); // Sleep 1/2 second
} catch (InterruptedException e) {
// Someone woke us up during sleep, that's OK
}
}
t.interrupt(); // Tell the thread to stop
t.join(); // Wait for the thread to cleanup and finish
That will give you resolution to about 1/2 second. By polling more often in the while loop, you can get that down.
Your runnable's run would look something like this:
public void run() {
while (true) {
try {
// Long running work
calculateMassOfUniverse();
} catch (InterruptedException e) {
// We were signaled, clean things up
cleanupStuff();
break; // Leave the loop, thread will exit
}
}
Update based on Dmitri's answer
Dmitri pointed out TimerTask, which would let you avoid the loop. You could just do the join call and the TimerTask you setup would take care of interrupting the thread. This would let you get more exact resolution without having to poll in a loop.
Depends on what the while loop is doing. If there is a chance that it will block for a long time, use TimerTask to schedule a task to set a stopExecution flag, and also .interrupt() your thread.
With just a time condition in the loop, it could sit there forever waiting for input or a lock (then again, may not be a problem for you).